The Most Famous Equation in Physics
E = mc^{2} is the most famous equation in physics; almost everyone with a reasonable education has heard of it. It describes the equivalence of energy and matter. Energy can be converted into matter, and matter can be converted into energy. This happens in nuclear reactions.
E = mc^{2}
energy = mass x (speed of light)^{2}
The speed of light is a constant, so its square is also going to be a constant – and a rather big number – so a large amount of energy is equivalent to a small amount of mass. The normal units used:
joules = kilograms x (metres per second)^{2}
J = kg·m^{2}·s^{2}
J = kg·m^{2}/s^{2}
Note that the kilogram is the only base unit which has a prefix. There are historical reasons for that.
If we want to convert energy to mass we rearrange the equation.
m = E / c^{2}
mass = energy / (speed of light)^{2}
How much energy does 1 kg equate to?
Let's use this equation to do a back of the envelope calculation to work out how much energy is in 1 kg.
E = mc^{2}
E = 1 x 300,000,000^{2}
= 9 x 10^{16} J
= 90 petajoules
To put that into some sort of perspective, 90 petajoules represents about 2.5 billion litres of petrol, or 21.5 megatonnes of TNT.
90 petajoules is just over the amount of electricity New Zealand produced from hydroelectric power stations in 2014.
How much mass converted to energy would it take to boil the world's ocean?
The average ocean surface temperature is 3.9 °C, because there's a lot of cold water deep in the oceans. Since the water surface is at sea level (!) it boils at 100 °C. Water takes about 4,180 J to heat 1 kilogram up by 1 °C (the specific heat) and 2,260,000 J to turn 1 kg of water at 100 °C into steam at 100 °C (the latent heat of vaporisation).
The average ocean depth is about 3,700 meters.
Earth's circumference (through the poles) is 40,000 km. That's because the metre was originally defined in 1793 as being one ten millionth the distance from the north pole to the equator (through Paris). The whole circumference is thus 40 million metres.
circumference = 2π x radius
r = c / 2π
Earth's radius = 40,000,000 / 2π
= 6,366,200 m
volume of a sphere = ^{4}/_{3}πr^{3}
volume of ocean = ^{4}/_{3}π(6.3662 x 10^{6})^{3 } ^{4}/_{3}π(6.3625 x 10^{6})^{}^{3}
= ^{4}/_{3}π( (6.3662 x 10^{6})^{3 } (6.36625 x 10^{6})^{}^{3} )
= 1.887 x 10^{18} m^{3}
= 1.887 x 10^{21} kg
energy to heat to 100°C = 4,180 x (100  3.9) x 1.887 x 10^{21}
= 7.58 x 10^{26} J
mass equivalent = 7.58 x 10^{26} / (3 x 10^{8})^{2}
= 8.4 x 10^{9} kg
= 8.4 billion kilograms
energy to evaporate = 2,260,000 x 1.887 x 10^{21}
= 4.26 x 10^{27} J
mass equivalent = 4.26 x 10^{27} / (3 x 10^{8})^{2}
= 4.73 x 10^{10} kg
= 47.3 billion kilograms
total mass equivalent = 8.4 billion + 47.3 billion
= 55.7 billion
So a total of 55.7 billion kilograms of mass converted to energy would be needed to heat and boil off the world's oceans. This is a lot of energy, and one reason why Earth is a nice place to live – all that water provides a huge stablising effect.
How much mass does the Sun convert into energy each second?
In the Sun four hydrogen atoms are converted (in several steps) into a single helium atom, but the helium atom does not have the same mass as the four hydrogen atoms it was formed from. Most of that lost mass has been converted to energy (a few weird particles are emitted as well).
We could just look up the solar luminance and plug that into E = mc^{2}, but let's back up a step or two and work it out from how much energy Earth receives each second. Wikipedia:
The Earth receives 174,000 terawatts (TW) of incoming solar radiation (insolation) at the upper atmosphere. Approximately 30% is reflected back to space while the rest is absorbed by clouds, oceans and land masses.
If we divide 174 petawatts, or 1.74 x 10^{17} W, by the crosssectional area of Earth, we'll get the power per square meter at one astronomical unit (1 AU, Earth's average orbit radius) from the Sun.
area of a circle = πr^{2}
Earth's crosssection area = π x 6,366,200^{2}
= 1.27 x 10^{14} m^{2}
solar irradiance = power / area
solar irradiance at 1 AU = 1.74 x 10^{17} W / 1.27 x 10^{14} m^{2}
= 1.37 kW/m^{2}
That's the amount of solar radiation at a range of 1 AU. We just need to multiply this by the surface area of a sphere with radius of 1 AU to get the power output of the Sun, and hence the energy it delivers each second (because 1 W = 1 J/s). 1 AU is about 150 million kilometres (150 billion metres).
surface area of sphere = 4πr^{2}
= 4π x 150,000,000,000^{2}
= 2.83 x 10^{23 }m^{2}
total power output of Sun = 1,370 W/^{}m^{2} x 2.83 x 10^{23 }m^{2}
= 3.87 x 10^{26} W*
That's 3.87 x 10^{26} joules each second. Assuming all of that comes from conversion of matter into energy, we can plug that value straight into the rearranged equation to get the mass the Sun converts into energy each second.
m = E / c^{2}
mass converted per second = 3.87 x 10^{26} / 3 x 10^{8}
= 4.3 x 10^{9} kg
4.3 billion kilograms is quite a lot, roughly the mass of 3 million cars at 1.4 tonnes each. (Japan makes almost 10 million vehicles per year.) Thankfully, the Sun is quite massive, so should last for quite a few seconds.
But what about...
OK, we can't wrap this up without doing just one more calculation.
time for the Sun to heat and boil oceans = 55.7 x 10^{9} / 4.3 x 10^{9}
= 13 seconds
If the total energy output of the Sun were put into heating and evaporating the world's oceans, it would take 13 seconds. That rate of heating would manifest as a really big explosion, since roughly 280 metres depth of water would be heated and evaporated each second. All the water would be exploded quickly enough to blow it off the planet and into space.
But wait, there's more. Because one side of the planet is mostly Pacific Ocean, Earth's orbit would be seriously affected. (I'm going to leave that calculation for another time.)
You would have a really bad day.
